A) 150 ppm
B) 600 ppm
C) 450 ppm
D) 300 ppm
Correct Answer: D
Solution :
[d] According to the law of gram equivalence \[mEq.HCl=mEq.Ca{{(HC{{O}_{3}})}_{2}}=mEq.(CaC{{O}_{3}})\] \[30ml\frac{N}{50}HCl=30ml\frac{N}{50}Ca{{(HC{{O}_{3}})}_{2}}=30ml\,of\frac{N}{50}CaC{{O}_{3}}\]\[=100\]ml of tap water. \[\therefore 30ml\frac{N}{50}HCl=\frac{wt.ofCaC{{O}_{3}}}{eq.wt.ofCaC{{o}_{3}}}\] Eq. wt. of \[CaC{{O}_{3}}=\frac{100}{2}=50\] \[\therefore \frac{30}{50\times 1000}=\frac{wt.ofCaC{{O}_{3}}}{50}\] \[\therefore wt.ofCaC{{O}_{3}}=\frac{30\times 50}{50\times 1000}=0.03g\] 100 ml of tap water has \[=0.03gofCaC{{O}_{3}}\] 1 ml of tap water \[=\frac{0.03}{100}\] \[{{16}^{6}}\]parts of tap water\[=\frac{0.03}{100}\times {{10}^{6}}\] Hardness=300 ppm
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