1.

Relative decrease in vapour pressure of an aqueous `NaCl` is `0.167`. Number of moles of `NaCl` present in `180g` of `H_(2)O` is:A. 2 molB. 1 molC. 3 molD. 4 mol

Answer» Correct Answer - B
RLVP=i(n)/(n+V)`
so `0.167=(2xxn)/(n+(180)/(18))`
so `n=1`


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