224ml `SO_2` at NTP react with 100ml, 0.1M NaOH and give non volatile product, which is dissolve in 36g of water, then vapour pressure of solution is `[X]xx10^(-2)` then value of X is
[Given `P^@_(H_2O) = 24mm of Hg at 25^@C`]

224 mL of SO_(2(g)) at 298 K and 1 atm is passed through 100 mL of 0.1 M NaOH solution. The non-volatile solute produced is dissolved in 36 g of water. The lowering of vapour pressure of solution (assuming the solution is dilute) (P_((H_2 O)) = 24 "mm of Hg) is" xx 10^(-2) mm of Hg, the value of x is _________. (Integer answer)

When 25 g of a non-volatile solute is dissolved in 100 g of water, the vapour pressure is lowered by 2.25xx10^(-1)mm of Hg. What is the molecular weight of the solute? [ P^(@)= 17.73 mm]

100ml solution of Na_3PO_4 contains 2.35g of Na^+ ion, then molarity of solution is [x] xx10^(-2) , then x is

When 25g of a non-volatile solute is dissolved in 100g of water, the vapour pressure is lowered by 0.225 mm. If the vapour pressure of water at 25^(@)C is 17.5 mm, what is the molecular mass of solute ?

The vapour pressure of CCl_4 at 25^@C is 143 mm Hg. If 0.5 gm of a non-volatile solute (mol.weight=65) is dissolved in 100g CCl_4 , the vapour pressure of the solution will be

In 100ml , 0.1N Na_2CO_3.xH_2O solution. Mass of solute is 1.43g, then value of X is: