Hardness of water is 200 ppm. Calculate the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) $200 \ ppm$ means $200 \ g$ of $CaCO_3$ in $10^6 \ g$ (or approximately $10^6 \ mL$) of water.$	ext{Molar mass of } CaCO_3 = 40 + 12 + (3 	imes

Vedclass · Accepted Answer

$2 	imes 10^{-3} \ M ; 4 	imes 10^{-3} \ N$

Vedclass · Answer

(A) $200 \ ppm$ means $200 \ g$ of $CaCO_3$ in $10^6 \ g$ (or approximately $10^6 \ mL$) of water.$	ext{Molar mass of } CaCO_3 = 40 + 12 + (3 	imes 16) = 100 \ g/mol$.$	ext{Molarity} = \frac{	ext{mass in } g}{	ext{Molar mass}} 	imes \frac{1000}{V_{mL}} = \frac{200}{100} 	imes \frac{1000}{10^6} = 2 	imes 10^{-3} \ M$.$	ext{Equivalent mass of } CaCO_3 = \frac{	ext{Molar mass}}{n	ext{-factor}} = \frac{100}{2} = 50 \ g/eq$.$	ext{Normality} = \frac{	ext{mass in } g}{	ext{Equivalent mass}} 	imes \frac{1000}{V_{mL}} = \frac{200}{50} 	imes \frac{1000}{10^6} = 4 	imes 10^{-3} \ N$.

Hardness of water is $200 \ ppm$. Calculate the molarity and normality of $CaCO_3$ in the water.

Similar Questions

At $NTP$,$40 \, mL$ of nitrogen reacts with hydrogen to form ammonia. According to the reaction $N_2 + 3H_2 \rightarrow 2NH_3$,the mass of ammonia formed will be ..... (in $, g$)

$A$ gaseous mixture contains $CH_4$ and $C_2H_6$ in equimolecular proportion. The weight of $2.24 \ L$ of this mixture at $S.T.P.$ is :- .............. $g$

$50 \ mL$ of $0.5 \ M$ oxalic acid is needed to neutralize $25 \ mL$ of sodium hydroxide solution. The amount in grams of $NaOH$ in $50 \ mL$ of the given sodium hydroxide solution is:

Volume of water needed to mix with $10 \ mL$ of $10 \ N$ $HNO_3$ to get $0.1 \ N$ $HNO_3$ is ........... $mL$.

$12 \ g$ of $Mg$ (atomic mass $24$) will react completely with acid to give:

Vedclass Test Series

Exam Paper Generator

Online Exam Module