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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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