Question

What is mole fraction of NaOH which is 200g and is in the solution of 1800g of water ​

Answer 1

Answer:

20/21

Explanation:

no. of moles= molecular mass/ given mass

moles of NaOH= 40/200=1/5

moles of water 18/1800=1/100

mole fraction=

number of moles of NaOH/number of moles of NaOH+number of moles of water

=1/5

/1/5 + 1/100

= 1/5*100/21

= 100/105

=20/21

Answer 2

Mole fraction is defined as the proportion of moles of that particular component(solute or solvent) in that solution by the sum of numbers of moles of all components in this solution.

$\sf Mole\:fraction\:of\:solvent (x1)=\frac{n1}{n1+n2}$

$\sf Mole\:fraction\:of\:solute (x2)=\frac{n2}{n1+n2}$

{here n1 =number of moles of solute, ( n2)=number of moles of solvent }

sum of mole fraction of solute and solvent is always 1

therefore,

$\sf x1+x2 = \frac{n1}{n1+n2}+\frac{n2}{n1+n2} = 1$

$\sf Moles\:of\:NaoH(n1) = \frac{Given\:molecular\:weight\:of\:NaoH}{Gram\:molecular\:weight\:of\:NaoH}$

$\sf =\frac{200}{40}$

$\sf =5\:mol$

Now, find n2

$\sf Moles\:of\:water (n2)=\frac{1800}{18}$

$\sf = 100\:mol$

Now ,we have to find the mole fractions,

$\sf Mole\:fraction\:of\:NaoH =x1=\frac{n1}{n1+n2}$

$\sf = \frac{5}{5+100}$

$\sf =\frac{5}{105}=\frac{1}{21}=0.4761$

$\sf Mole\:fraction\:of\:Water (x2)=\frac{n2}{n1+n2}$

$=\frac{100}{5+100}$

$=\frac{100}{105}=0.9523$