6 g of a non volatil, non electrolyte X dissolved in 100g of water freezes at `-0.93^@C`. The molar mass of X in g `mol^(-1)` is `(K_f of H_2O=1.86kgmol^(-1))`.

6g of non-volatile, non-electrolyte x dissolved in 100g of water freezes -0.93^(@)C . The molar mass of x in "g mol"^(-1) is : (K_(f) "of" H_(2)O=1.86 "K kg mol"^(-1))

6g of non-volatile, non-electrolyte x dissolved in 100g of water freezes -0.93^(@)C . The molar mass of x in "g mol"^(-1) is : (K_(f) "of" H_(2)O=1.86 ^(@)C kg mol"^(-1))

6 grams of a substance 'A' dissolved in 100 g of water freezes at - 0.93^(@)C . The molecular mass of 'A' is ( K_(f)=1.86 Km^(-1) ).

A solution containing 18g of non - volatile non - electrolyte solute is dissolved in 200g of water freezes at 272.07K. Calculate the molecular mass of solute. Given K_(f)=1.86kg//mol and freezing point of water = 273K

If 10.0 g of a non-electrolyte dissolved in 100 g of water lowers the freezing point of water by 1.86^(@)C , the molar mass of the non-electrolyte is ( K_(f)=1.86 Km^(-1) )