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Grade 12Algebra

The vapour pressure of pure water at 30°C is 31.80 mm of hg. How many grams of urea(molecular mass =60) should be dissolved in 1000 g of water to lower the vapour pressure by 0.25 mm of hg?

khushbu

9 Years agoGrade 12

3 Answers

Manan patidar

8 Years ago

The vapour pressure of pure water at 30°C is 31.80 mm of hg. How many grams of urea(molecular mass =60) should be dissolved in 1000 g of water to lower the vapour pressure by 0.25 mm of hg?

Answer is 26.2g

Arun

8 Years ago

Dear student

As we know

p° - p /p° = W2 M1 /M2 W1

Now put the values

0.25 /31.8 = W2 *18 /(60* 100)

On calculation,

W2 = 2.62 gm

Regards

Arun (askIITians forum expert)

Laishngbam karoke

5 Years ago

Given

Urea (A), WA= ?

MA= 60g/mol

Water (B), WB = 100g

MB = 18g/mol

P° = 31.80mm

P°- Ps = 0.25mm

Now,

P° - Ps /P° = WA /MA× MB / WB

0.25/31.80= WA /60× 18 /100

0.25/31.80= WA/30 × 9/100

WA = 0.25×30× 100/31.80× 9

= 0.25× 10 × 100/31.80× 3

= 250/31.8 ×3

= 2500/954

WA = 26.2 g

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