Question

18 gm glucose is present in 90 gm of water. Calculate the mole fraction of glucose in solution

Answer 1

Answer:

So the number of moles of glucose present is 90180=0.5 (180 is the relative molecular mass of glucose) and the number of moles of water is 18018=10 (18 is the relative molecular mass of water)

Answer 2

Explanation:

Molecular mass of water= 2×1+1×16=18g

=2×1+1×16=18g

For 90g178.2g water n

A

=5

Molecular mass of glucose 6×12+12×1+6×16=180g

For 18g glucose n

B

=0.1

Now, X

B

=

0.1+5

0.1

=0.0196 ~0.02

Answer 3

Explanation:

Molecular mass of water= 2×1+1×16=18g

=2×1+1×16=18g

For 90g178.2g water n

A

=5

Molecular mass of glucose 6×12+12×1+6×16=180g

For 18g glucose n

B

=0.1

Now, X

B

=

0.1+5

0.1

=0.0196 ~0.02

Answer 4

Explanation:

Molecular mass of water= 2×1+1×16=18g

=2×1+1×16=18g

For 90g178.2g water n

A

=5

Molecular mass of glucose 6×12+12×1+6×16=180g

For 18g glucose n

B

=0.1

Now, X

B

=

0.1+5

0.1

=0.0196 ~0.02

Answer 5

Explanation:

Molecular mass of water= 2×1+1×16=18g

=2×1+1×16=18g

For 90g178.2g water n

A

=5

Molecular mass of glucose 6×12+12×1+6×16=180g

For 18g glucose n

B

=0.1

Now, X

B

=

0.1+5

0.1

=0.0196 ~0.02