Calculate the normality of 1.11 gm of C a C l 2 is dissolved in 250 ml solution. Options: A . &nbsp 0.1 N B . &nbsp 0.25 N C . &nbsp 0.08 N D . &nbsp 0.3 N

Answer: Option C : C Normality = N o . o f g r a m e q u i v a l e n t s o f s o l u t e V o l u m e o f s o l u t i o n ( i n l i t r e ) equivalent mass = M o l e c u l a r M a s s V a l e n e c y f a c t o r = 111 2 [ C a + 0 → C a 2 + + 2 e − ] = 55.5 No of gram equivalents = 1.11 55.5 ⇒ Normality = 1.11 55.5 250 1000 = 0.08 N

Calculate the normality of 1.11 gm of CaCl2 is dissolved in 250 ml soluti

Discussion Forum : Mole Concept

Question -

Calculate the normality of 1.11 gm of $C a C l_{2}$ is dissolved in 250 ml solution.

Options:

A . 0.1 N

B . 0.25 N

C . 0.08 N

D . 0.3 N

Answer: Option C
:
C

Normality = $\frac{N o . o f g r a m e q u i v a l e n t s o f s o l u t e}{V o l u m e o f s o l u t i o n (i n l i t r e)}$

equivalent mass = $\frac{M o l e c u l a r M a s s}{V a l e n e c y f a c t o r}$

= $\frac{111}{2}$ [ ${C a}^{+ 0}$ $\to$ ${C a}^{2 +}$ + $2 e -$ ] = 55.5

No of gram equivalents = $\frac{1.11}{55.5}$

$\Rightarrow$ Normality = $\frac{\frac{1.11}{55.5}}{\frac{250}{1000}}$ = 0.08 N

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