22-04-2018 /study/ppm to kg/ha, meq/100g to ppm, mg/l

Suppose Mr. K. has 10 plots each measuring 3m x 3m for planting stylosanthes. If the Optimal Fertilizer requirement for stylosanthes is 100 kg N ha-1. Calculate the amount of NPK 20-20-20 Fertilizer that will be required for each plot and the total land area. Express your answer in **kg ha-1**, **ppm**,** mgkg-1****mgL-1** and **Mtha-1**

One hectare of stylosanthes requires 180 kg of N.

NPK 20-20-20 fertilizer has 20% of N.

100kg of NPK 20-20-20 contains 20kg of N.

To solve for the amount of NPK 20-20-20 that will give 180kg of N.

100kg of NPK 20-20-20 / Ykg of NPK 20-20-20= 20kg of N / 180kg of N.

Ykg x 20kg = 100kg x 180kg

Y = (18000 / 20) kg of NPK 20-20-20.

Y= 900kg of NPK 20-20-20.

THEREFORE 900kg of NPK 20-20-20 will be required for one hectare of stylosanthes.

If 900kg of NPK 20-20-20 is required for 1 HECTARE = 10 000 m2 of stylosanthes.

Ykg of NPK 20-20-20 iwill be required for 3m x 3m = 9 m2 of stylosanthes.

90kg of NPK 20-20-20 / Ykg of NPK 20-20-20 = 10 000 m2 / 9m2.

10 000 x Y = 90 X 9

Y= (810/10 000) kg of NPK 20-20-20 = 0.081kg of NPK or 81g of NPK 20-20-20

THEREFORE 0.081kg of NPK 20-20-20 will be required for a 3m x 3m plot of stylosanthes.

For Hectare

90kg N ha--1

Y ibs/acre = ( 1.12 x Y) kg/ha

100.8 ibs/acre = 90kg/ha x 1.12

Y ibs/acre = ( Y / 2 ) ppm

100.8 ibs/acre= (100.8 / 2) ppm = 50.4 ppm.

Y ppm = Y mg/kg

50.4ppm = 50.4mgkg-1

Y ppm = Y mg/litre

50.4ppm = 50.4mgL-1

Y kg/ha = ( Y / 1 000 ) metric tonnes /ha

90kg/ha = (90 / 1 000) Mtha-1 = 0.09Mtha-1

For 9m2 Plot

0.081kg N per plot

Y ibs/acre = ( 1.12 x Y) kg/ha

0.09072 ibs/acre = 0.081kgha-1 x 1.12

Y ibs/acre = ( Y / 2 ) ppm

0.09072 ibs/acre= (0.09072 / 2) ppm = 0.04536 ppm.

Y ppm = Y mg/kg

0.04536 ppm= 0.04536 mgkg-1

Y ppm = Y mg/litre

0.04536 ppm = 0.04536 mgL-1

Y kg/ha = ( Y / 1 000 ) metric tonnes /ha

0.04536 kg/ha = 0.04536 / 1 000) Mtha-1 = 0.00004536 Mtha-1

see all soil conversion units (ppm: meq/100g: cmol/kg e.t.c)here

Sample Soil Fertility Question 2:

Suppose Ammonium Sulphate is to be applied to a 100 m2 plot of maize planted at 75 cm apart and 25 cm within the row at the rate of 300kg N ha-1 Calculate the amount of fertilizer to be applied

(i)to the whole plot (ii)per row (iii)per each stand of maize.

If 200kg of N is to be applied per hectare or 10 000 m2,

then Y kg of N will be applied per 100 m2

200kg of N / Ykg of N = 10 000 m2 / 100 m2.

Y x 10 000 = 200 x 100 (Through Cross Multiplication)

Y = 20 000 / 10 000

Y= 2kg of N.

Therefore 2 kg of N will be applied per 100 m2.

(i) The amount of Ammonium Sulphate Fertilizer needed per 100 m2 plot = The amount of ammonium sulphate fertilizer that will supply 2 kg of N.

Ammonium Sulphate Fertilizer (NH4)2SO4 contains 21 % of N.

It means 100kg of (NH4)2SO4 will supply 21 kg of N.

Therefore Y kg of (NH4)2SO4 will supply 2 kg of N.

100kg of (NH4)2SO4 / Ykg of (NH4)2SO4 = 21 kg of N / 2 kg of N.

Y x 21 = 100 x 2 (Through Cross Multiplication)

Y = 200 / 21

Y= 9.52 kg of (NH4)2SO4.

Therefore 9.52 kg of Ammonium Sulphate is the amount needed to supply 2 kg of N for 100 m2 plot .

9.52 kg of Ammonium Sulphate is also the amount needed to supply N to a 100 m2 plot at the rate of 200kg of N per hectare.

(ii)

If 9.52 kg of Ammonium Sulphate is to be applied per 100 m2 plot,

A row of the plot will measure 10m in length but 75cm in width since they are 75cm apart.

Therefore the area of a row is 10 x 0.75 = 7.5 m2.

then Y kg of Ammonium Sulphate will be applied per row = 7.5 m2.

9.52 kg of Ammonium Sulphate / Ykg of Ammonium Sulphate = 100 m2 / 7.5 m2.

Y x 100 = 7.5 x 9.52 (Through Cross Multiplication)

Y = 71.4 / 100

Y= 0.714 kg of Ammonium Sulphate per row or 714g of Ammonium Sulphate per row.

(iii)

If 9.52 kg of Ammonium Sulphate is to be applied per 100 m2 plot,

The number of maize stand per the 100 m2 plot will equal 100 m2 / (0.25m x 0.75m) since they are spaced at 75cm x 25cm.

Therefore the number of maize stands is 100 m2 / 0.1875m2 = 533.33

it means 9.52kg of Ammonium Sulphate is applied to 533.33 stands of maize.

Y kg of Ammonium Sulphate is applied to a stand of maize.

9.52 / Y = 533.33 / 1

Y x 533.33 = 9.52

Y = 9.52 / 533.33

Y= 0.0179kg of Ammonium Sulphate Fertilizer.

Therefore 0.0179kg of Ammonium Sulphate Fertilizer is the amount to apply per each stand of maize.

see all soil conversion units (ppm: meq/100g: cmol/kg e.t.c)here

Sample Soil Fertility Question 3:

Mr. Fadama has a plantation of *Corchorus Olitorius* containing 15 plots each of which is measured 4m X 3m.

If 60 kg N ha-1 is required by *Corchorus Olitorius* which should be applied 30 kg N ha-1as Urea and 30 kg N ha-1 as Organic Fertilizer (which contains 3.5% of N)

Calculate the amount of Urea and Organic Fertilizers needed for:

(i) each plot (ii) 20 plots (iii) an hectare.

If 60kg of N is to be applied per hectare or 10 000 m2,

then Y kg of N will be applied per (3 x 4) m2 = 12 m2,

60kg of N / Ykg of N = 10 000 m2 / 12 m2.

Y x 10 000 = 60 x 12 (Through Cross Multiplication)

Y = 720 / 10 000

Y= 0.072 kg of N.

Therefore 0.072 kg of N will be applied per 12 m2.

(A) Urea

(i) The amount of Urea Fertilizer needed per 12 m2 plot = The amount of Urea fertilizer that will supply 0.072 / 2 kg of N.(since Urea will only supply half (30 kg N ha-1))

Urea Fertilizer contains 46 % of N.

It means 100kg of Urea will supply 46 kg of N.

Therefore Y kg of Urea will supply 0.036 kg of N.

100kg of Urea / Ykg of Urea = 46 kg of N / 0.036 kg of N.

Y x 46 = 100 x 0.036 (Through Cross Multiplication)

Y = 3.6 / 46

Y= 0.0783 kg of Urea.

Therefore 0.0783 kg of Urea.is the amount needed to supply 0.036 kg of N for 12 m2 plot .

0.0783 kg of Urea is also the amount needed to supply N to a 12 m2 plot at the rate of 30kg of N per hectare.

(ii)

since 0.0783 kg of Urea is needed per a 12m2plot, therefore 15 x 0.0783 kg of urea is needed for 15 plots.

1.1745 kg or 1174.5 g of urea is needed for 15 plots .

If 0.0783 kg of Urea is to be applied per 12 m2plot,

then Y kg of Urea will be applied per 10 000 m2 = 1 hectare,

0.0783 kg of Urea / Ykg of Urea = 12 m2 / 10 000 m2.

0.0783 x 10 000 = Y x 12

Y = 783 / 12

Y = 65.25 kg of Urea.

Therefore 65.25 kg of Urea will required to suppy 30 kg N ha-1

(B) Organic Fertilizer.

(i) The amount of Organic Fertilizer needed per 12 m2 plot = The amount of Urea fertilizer that will supply 0.072 / 2 kg of N.(since Organic fertilizer will only supply half (30 kg N ha-1))

Organic Fertilizer contains 3.5 % of N.

It means 100kg of Organic Fertilizer will supply 3.5 kg of N.

Therefore Y kg of Organic Fertilizer will supply 0.036 kg of N.

100kg of Organic Fertilizer / Ykg of Organic Fertilizer = 3.5 kg of N / 0.036 kg of N.

Y x 3.5 = 100 x 0.036 (Through Cross Multiplication)

Y = 3.6 / 3.5

Y= 1.03 kg of Organic Fertilizer.

Therefore 1.03 kg of Organic Fertilizer.is the amount needed to supply 0.036 kg of N for 12 m2 plot .

1.03 kg of Organic Fertilizer is also the amount needed to supply N to a 12 m2 plot at the rate of 30kg of N per hectare.

(ii)

since 1.03 kg of Urea is needed per a 12m2plot, therefore 15 x 1.03 kg of urea is needed for 15 plots.

15.45 kg or 15450 g of urea is needed for 15 plots .

(iii)

If 15.45 kg of Urea is to be applied per 12 m2plot,

then Y kg of Urea will be applied per 10 000 m2 = 1 hectare,

15.45 kg of Urea / Ykg of Urea = 12 m2 / 10 000 m2.

15.45 x 10 000 = Y x 12

Y = 154500 / 12

Y =12875 kg of Urea.

Therefore 12875 kg of Urea will required to suppy 30 kg N ha-1

see all soil conversion units (ppm: meq/100g: cmol/kg e.t.c)here

Sample Soil Fertility Question 4:

Given 2.4 meq Ca per 100g of soil (i.e 2.4 cmol / kg)(i) Calculate the concentration of Ca in that soil in (i) ppm (ii) pound per acre (iii) kg per ha.

meq/100g of soil = cmol /kg.

Y meq/100g = ( ( equivalent weight x 10 x Y ) ) ppm.

molar mass of Ca = 40.

equivalent weight of Ca = 40/2 = 20

2.4 meq/100g =(20 x 10 x 2.4) ppm.

2.4 meq/100g = 480 ppm

(ii)

pounds per acre

Y ppm = ( 2 x Y ) ibs/acre ( pounds/acre )

480 ppm = (2 x 480) ibs/acre ( pounds/acre ).

480 ppm = 960 pounds/acre.

(iii)

kg per ha

Y ibs/acre = ( 1.12 x Y) kg/ha

960 pounds per acre = (1.12 x 960) kg / ha.

960 pounds per acre = 1075.2 kg/ha

Sample Soil Fertility Question 5:

Suppose a farmer has tomato plantation of 200 x 100 m2 and tomato requires 128 mg K kg-1 . Calculate the amount of fertilizer required per hectare of land and per the land area. The available fertilizer is muriate of potash(MOP). M.O.P. contains 50% of K. What is the amount of fertilizer required per row, if the spacing is 50cm x 50cm. (ii) What amount of fertilizer would be applied per plant stand?

128 mg K kg-1 = 128 ppm

Y ppm = ( 2 x Y ) ibs/acre ( pounds/acre )

128 ppm = (2 x 128) ibs/acre. = 256 pounds/acre.

Y ibs/acre = ( 1.12 x Y) kg/ha

256 pounds/acre = (1.12 x 256) kg/ha.

256 ibs/acre = 286.72 kg/ha.

MOP contains 50% of K . it means 100 kg of MOP contains 50 kg of K.

Y kg of MOP will supply 286.72 kg of K

100 kg of MOP / Y kg of MOP = 50 kg of K / 286.72 kg of K

Y x 50 = 100 x 286.72

Y = 28672 / 50

Y = 573.44 kg of MOP

The amount of fertilizer (MOP) required per hetare of land =573.44 kg of MOP.

(ii).

if 573.44 kg of MOP is required per 1 hectare or 10 000 m2

then Y kg of MOP will be required per 200 x 100 m2 = 20 000m2

573.44 kg of MOP / Y kg of MOP = 10 000 m2/ 20 000 m2

Y x 10 000 = 573.44 x 20 000

Y = (573.44 x 20 000) / 10 000

Y = 573.44 X 2 = 1146.88

Y = 1146.88 kg of MOP

The amount of fertilizer (MOP) required per the land area = 1146.88 kg of MOP.

(iii)

Per row

spacing = 50cm x 50cm. It means the spacing between rows is 50cm.

It means the rows lengths are 100m but their widths are 50 cm since it is the spacing between rows.

the area of the rows will now be 100m x 0.5m = 50 m2.

therefore if 10 000 m2 of land requires 573.44 kg of MOP

50 m2 of land will require ((50 x 573.44) / 10 000) kg of MOP =2.8672 kg of MOP .

2.8672 kg of MOP will be required per row.

(iv)

Since spacing is 50 cm X 50 cm = 0.5m x 0.5m = 0.25 m2

Therefore the number of tomato stands on the plot is (200 x 100) m2 / 0.25 m2 = 20 000 / 0.25 = 80 000 stands.

Since 80 000 stands require 1146.88 kg of MOP.

Therefore 1 stand will require Y kg of MOP.

Y = 1146.88 / 80 000 kg of MOP = 0.014336 kg of MOP.

Therefore 0.014 kg of MOP will be required per stand.

see all soil conversion units (ppm: meq/100g: cmol/kg e.t.c)here

Sample Soil Fertility Question 6:

If 10g of soil was extracted with 100 ml 0.1N ammonium acetate. The extract solution contained 0.008% K. what is the concentraton of K in the soil in cmol/kg? EXpress your answer in ppm and kg/ha soil.

Dilution factor ( no unit ) = volume of extracting solution / mass of the soil.

dil factor = 100 ml of 0.1N ammonium acetate / 10 g of soil.

dil factor = 100 /10 = 10

K concentration in the soil = dil. factor x final concentration = 10 x 0.008 % = 0.08 %.

y % = ( y x 10 000) ppm

0.08 % = (0.08 x 10 000) ppm = 800 ppm

Y ppm = ( 1.12 x 2 x Y) kg/ha

800ppm = (1.12 x 2 x 800) kg/ha = 1 792 kg/ha

see all soil conversion units (ppm: meq/100g: cmol/kg e.t.c)here

Given the following soil text data, answer the questions that follow

Nutrient | meq/100g |

Ca2+ | 6.2 |

K+ | 2.0 |

H2PO4 | 1.8 |

SO42- | 4.0 |

Mg2+ | 1.8 |

Na+ | 0.05 |

Al3+ | 8.4 |

NH4+ | 0.4 |

a) Calculate the CEC of the soil in cmolkg-1

b) Calculate the percent base saturation of the soil.

c) Calculate the amount of potassium in the soil in ppm and kgha-1.

d) Calculate the amount of calcium in the soil in ppm and kgha-1.

e) Calculate the percent aluminium saturation in the soil.

1) CEC = SUM of the meq/100g of all the cations = ( Ca2+, + K+ + Mg2+ + Na+ + Al3+ + NH4+)meq/100g = (6.2+2.0+1.8+0.05+8.4+0.4)meq/100g = 18.85 meq/100g

b) percent base saturation = (SUM of basic cations / ECEC (sum of acidic and basic cations) ) x 100

sum of basic cations. = (( Ca2+, + K+ + Mg2+ + Na+ + NH4+)meq/100g = (6.2+2.0+1.8+0.05+0.4)meq/100g = 10.45meq/100g).

sum of acidic and basic cations = (( Ca2+, + K++ Mg2+ + Na+ + Al3+ + NH4+)meq/100g = (6.2+2.0+1.8+0.05+8.4+0.4)meq/100g = 18.85 meq/100g).

percent base saturation = (10.45 / 18.85) x 100 = 55.4%

c) amount of potassium saturation in the soil in ppm and kg/ha

K+ in cmol/kg = 2.0 cmol/kg

equivalent weight of K+ = 39/1 = 39

Y meq/100g = ( ( equivalent weight x 10 x Y ) ) ppm

cmol/kg=meq/100g

2.0 cmol/kg= ( ( 39 x 10 x 2.0 ) ) ppm

2.0 cmol/kg= 780ppm

Yppm= (2 x 1.12 x Y) kg/ha

780 ppm = 2 x 1.12 x 780 kg/ha = 1 747.2kg/ha

c) amount of calcium saturation in the soil in ppm and kg/ha

Ca+ in cmol/kg = 6.2 cmol/kg

equivalent weight of Ca+ = 40/2 = 20

Y meq/100g = ( ( equivalent weight x 10 x Y ) ) ppm

cmol/kg=meq/100g

6.2 cmol/kg= ( ( 40 x 10 x 6.2 ) ) ppm

6.2 cmol/kg= 2 480 ppm

Yppm= (2 x 1.12 x Y) kg/ha

780 ppm = (2 x 1.12 x 2 480 kg/ha) = 5 555.2kg/ha

d) percent aluminium saturation = (exchangeable aluminium x 100) / ECEC(SUM of basic and acidic cations)

percent aluminium saturation = (8.4 x 100) / 18.85 = 44.56%

Suppose Mr. Tamedo carried out a field research on a plantation of 1/4 hectare. The field was then laid out at 4m x 3m per plot. If 10g of soil sample taken from the field was extracted with 100ml amonium acetate. 5ml of the filtrate was made up to 50ml. Potassium concentrationof the final solution was 0.08ppm . If NPK 10-10-10 fertilizer was applied to supply potassium at 20kg ha-1, Nitrogen, and Phosphorus. Calculate (i) the total dilution factor. (ii) Concentration of K in the soil in (a) ppm (b) mgkg-1 (c) mgL-1 (d) meq/100g of soil (e) cmolkg-1 (f) kgha-1 (g) tonha-1. (iii) Calculate the total amount of NPK required for the whole plantation. (iv) Calculate the amount of N and P that will be supplied to each plot by the amount of NPK in iii.

(i) Dilution factor ( no unit ) = volume of extracting solution / mass of the soil

Dilution factor = final volume / initial volume.

The first dilution factor = 100 / 10 = 10

The second dilution factor= 50 / 5 = 10.

The total dilution factor = the first dilution factor x the second dilution factor.

The total dilution factor = 10 x 10 =100

(ii) Concentration of K in the soil = dilution factor x concetration of the final solution.

(a) K conc. = 100 x 0.08ppm = 8ppm.

(b)

Y ppm = Y mg/kg

8ppm = 8 mgkg-1

(c) Y ppm = Y mg/L

8ppm = 8 mgL-1

(d)

Y ppm = ( ( Y x 0.1) / equivalent weight) meq/100g (milliequivalent/100g)

equivalent weight of K = 39/1 = 39

8 ppm = (8 x 0.1) / 39 meq/100g = 0.0205 meq/100g

(e) Y meq/100g = Y cmolkg-1

0.0205 meq/100g = 0.0205 cmollkg-1

(f) Y ppm = (2 x 1.12 x Y) kgha-1

8 ppm = (2 x 1.12 x 8) kgha-1= 17.92 kgha-1

(g) tonnha-1

Y kg/ha = ( Y / 1 000 ) tonnes /ha

17.92 kg/ha = (17.92 / 1000) tonnha-1

17.92 kg/ha = 0.1792 tonnha-1

by SODIQ

he is a seasoned educationist. He obtained his bachelor degree from Obafemi Awolowo University. He has written many educational articles on different blogs

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