What is the degree of hardness of a sample of water containing 24 mg of $MgS{O_4}$ (molecular mass 120) per kg of water?
A.10 ppm
B.15 ppm
C.20 ppm
D.25 ppm
Answer
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Hint:Here, the hardness of water is permanent. The hardness of water is calculated in terms of the weight of $CaC{O_3}$. 1ppm = 1mg/L. The molecular weight of $CaC{O_3}$ = 100. The degree of hardness is expressed as parts per million (ppm).
Complete step by step answer:
The hardness of water is of two types: permanent and temporary.
Temporary hardness is due to the presence of magnesium and calcium hydrogen carbonates.
Presence of insoluble salts like $MgS{O_4}$ causes permanent hardness of the water.
The amount of hardness causing substances in a certain volume of water measures the degree of hardness.
The hardness of water is always calculated in terms of calcium carbonate ($CaC{O_3}$). However, it is never responsible for causing hardness as it is insoluble in water.
.
As mentioned, the degree of hardness is expressed as parts per million (ppm) and thus may be defined as the number of parts by weight of $CaC{O_3}$ (equivalent to Mg salts) present in a million (${10^6}$) parts by weight of water.
It is given that 1 kg of water contains 24 mg of $MgS{O_4}$.
i.e., 1000 g of water contains 24 mg of $MgS{O_4}$
therefore, 1 million grams of water (${10^6}$g ) contains 24 mg $ \times $ 1000 = 24 g of $MgS{O_4}$
1 mole of $MgS{O_4}$ = 1 mole of $CaC{O_3}$
The molecular weight of $MgS{O_4}$ = 120g/mol and $CaC{O_3}$= 100g/mol
120 g of $MgS{O_4}$ = 100 g of$CaC{O_3}$
Therefore, the equivalent amount of $CaC{O_3}$ for 24 mg of $MgS{O_4}$ = 24 $ \times $ 100 /120 = 20 ppm.
Hence, the degree of hardness of water is 20 ppm.
The correct answer is (C) 20 ppm.
Note: Here, 1 ppm = 1 part of $CaC{O_3}$ in ${10^6}$ parts of water
-$CaC{O_3}$ is chosen as the standard for calculating the hardness of water because its molecular weight is 100 and this makes it easier to do the calculations.
-Calcium carbonate is not responsible for the hardness of the water.
-Another method used to detect the hardness of water is by performing complexometric titrations using EDTA.
Complete step by step answer:
The hardness of water is of two types: permanent and temporary.
Temporary hardness is due to the presence of magnesium and calcium hydrogen carbonates.
Presence of insoluble salts like $MgS{O_4}$ causes permanent hardness of the water.
The amount of hardness causing substances in a certain volume of water measures the degree of hardness.
The hardness of water is always calculated in terms of calcium carbonate ($CaC{O_3}$). However, it is never responsible for causing hardness as it is insoluble in water.
.
As mentioned, the degree of hardness is expressed as parts per million (ppm) and thus may be defined as the number of parts by weight of $CaC{O_3}$ (equivalent to Mg salts) present in a million (${10^6}$) parts by weight of water.
It is given that 1 kg of water contains 24 mg of $MgS{O_4}$.
i.e., 1000 g of water contains 24 mg of $MgS{O_4}$
therefore, 1 million grams of water (${10^6}$g ) contains 24 mg $ \times $ 1000 = 24 g of $MgS{O_4}$
1 mole of $MgS{O_4}$ = 1 mole of $CaC{O_3}$
The molecular weight of $MgS{O_4}$ = 120g/mol and $CaC{O_3}$= 100g/mol
120 g of $MgS{O_4}$ = 100 g of$CaC{O_3}$
Therefore, the equivalent amount of $CaC{O_3}$ for 24 mg of $MgS{O_4}$ = 24 $ \times $ 100 /120 = 20 ppm.
Hence, the degree of hardness of water is 20 ppm.
The correct answer is (C) 20 ppm.
Note: Here, 1 ppm = 1 part of $CaC{O_3}$ in ${10^6}$ parts of water
-$CaC{O_3}$ is chosen as the standard for calculating the hardness of water because its molecular weight is 100 and this makes it easier to do the calculations.
-Calcium carbonate is not responsible for the hardness of the water.
-Another method used to detect the hardness of water is by performing complexometric titrations using EDTA.
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