Chemistry, asked by vaidehisathe8319, 10 months ago

What weight of solute in grams (molecular weight ) is required to dissolve in of water to reduce the vapour pressure by?

Answers

Answered by swan030782
0

Answer:

2M

Explanation:

Let the weight of solute (w2) = w grams

Weight of solvent (w1)= 180 grams

Molar mass of solvent H2O(M1) = 18

Moalr mass of solute (M2) = M

Let the vapour pressure of pure solvent (p1°) = x

Vapour pressure of solution (p1) = (x*4)/5

(p1°- p1)/ p1°= (w2*M1)/(w1*M2)

(x- 4x/5)x = (w *18)/(180*M2)

(1-4/5) = w/(10*M)

w=1/5*10*M

w= 2M

Therefore weight of the solute = 2M

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