Question

If 0.500L solution weighs 596g and contains 90.0g of oxalis acid, the concentration can be expressed as ?M, or ?(m/m)%, or ?(m/v)%.

Answer 1

Answer : The concentration can be expressed as, 1.99 M, or 15.1(m/m)%, or 18(m/v)%

Solution : Given,

Volume of solution = 0.5 L = 500 ml        (1 L = 1000 ml)

Mass of solution = 596 g

Mass of oxalic acid, (solute) = 90 g

Molar mass of oxalic acid = 90.03 g/mole

First we have to calculate the concentration in terms of M(molarity).

Molarity : Molarity is defined as the number of moles of solute present in one liter of solution.

Formula used : $M=\frac{w_{solute}}{M_{solute}\times V_s}$

where,

M = molarity of solution

$w_{solute}$ = mass of solute(oxalic acid)

$M_{solute}$ = molar mass of solute(oxalic acid)

$V_s$ = volume of solution(in liters)

Now put all the given values in this formula, we get

$M=\frac{90g}{90.03g/mole\times 0.5L}=1.99mole/L=1.99M$

Now we have to calculate the concentration in terms of (m/m)%

(m/m)% : It is defined as the mass of solute present in mass of solution in grams.

Formula used : $(m/m)\%=\frac{w_{solute}}{w_s}\times 100$

where,

$w_s$ = mass of solution

$(m/m)\%=\frac{90g}{596g}\times 100=15.1%$

Now we have to calculate the concentration in terms of (m/v)%

(m/v)% : It is defined as the mass of solute present in one milliliter of solution.

Formula used : $(m/v)\%=\frac{w_{solute}}{V_s}\times 100$

where,

$w_s$ = mass of solution

$(m/m)\%=\frac{90g}{500ml}\times 100=18%$