A beaker containing $20$ gm sugar in $100$ gm water and another containing $10$ gm sugar in $10$ gm water are placed under a bell- jar and allowed to stand until equilibrium is reached. How much water will be transferred from one beaker to other?

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Solution

Molarity of $10 g$ sugar in $100 m l$ water.
molarity $= (\frac{10}{342}) (\frac{1000}{100}) = 0.2932 M$
molarity of $20 g$ sugar in $100 m l$ water
molaity $(\frac{20}{342}) (\frac{1000}{100}) = 0.5840 M$
now $H_{1} V_{1} = M_{2} V$
So to make equal concentration equal amount of volume
i.e., $10 m l$ must be transferred out to other beaker

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A beaker containing 20gm sugar in 100 gm water and another containing 10 gm sugar in 10 gm water are placed under a bell- jar and allowed to stand until equilibrium is reached. How much water will be transferred from one beaker to other?

Molarity of 10g sugar in 100 ml water.molarity=(10342)(1000100)=0.2932Mmolarity of 20g sugar in 100 ml water molaity (20342)(1000100)=0.5840M now H1V1=M2VSo to make equal concentration equal amount of volume i.e., 10 ml must be transferred out to other beaker

A beaker containing $20$ gm sugar in $100$ gm water and another containing $10$ gm sugar in $10$ gm water are placed under a bell- jar and allowed to stand until equilibrium is reached. How much water will be transferred from one beaker to other?