A) 0.3
B) 1.0
C) 3.0
D) 0.1
Correct Answer: A
Solution :
Given that, volume = 500 mL = 0.5 litre weight of solute \[(\omega )=4.9\text{ }g\]equivalent weight of \[{{H}_{3}}P{{O}_{4}}=\frac{molecular\text{ }weight}{basicity}\] \[=\frac{3\times 1+31+4\times 16}{3}=\frac{98}{3}\] Hence, normality of solution (N) \[=\frac{\omega }{equivalent\text{ }weight\times volume}\] \[=\frac{4.9}{\frac{98}{3}\times 0.5}=0.3\,N\]
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