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:

Relative decrease in vapour pressure in vapour pressure of an aqueous solution NaCI is 0.1067, Number of moles of NaCI present in 180 g H_(2)O is:

Relative lowering in vapour pressure of an aqueous NaCl solution is 1/6. What is the number of moles of NaCl present in 180g H_(2)O

Relative decrease in vapour pressure of an aqueous solution conataining 2 mol [Cu(NH_(3))_(3)Cl] Cl in 3 mol H_(2) is (1)/(2) . When the given solution reation with excess fo AgNO_(3) solution the number moles of AgCl produced is

The relative decreases in the vapour pressure of an aqueous solution containing 2 mol [Cu(NH_(3)_(3)Cl] in 3 mol H_(2)O is 0.50 . On reaction with AgNO_(3) , this solution will form

The number of moles of NaCl in 250cm^(3) of 0.50 M NaCl is

Calculate relative lowering of vapour pressure of 0.161 molal aqueous solution