Question

The strngth of 10^-2M Na2CO3 solution in terms of molality will be (density of solution =1.1 g /ml )

Answer 1

Answer:

The strngth of 10^-2M Na2CO3 solution in terms of molality is 9×10^(-3) moles

Explanation:

Here 10^(-2)M Na2CO3

Molar Mass of Na2CO3 is (46+12+48)= 106 gm

Now, 10^(-2) M = 10^(-2) Molar = 10^(-2) Molar of Na2CO3= 10^(-2) Moles of NA2CO3 in 1000 ml solution

Molality=moles of solute in 1kg of solution

So, S =(1.1/ml)= (m/v), here v= 1000ml and M=1100 gm

We know that, Mass of solvent= Mass of solution - Mass of solute

Mole of Na2CO3= 106 gm and 10^(-2) moles= ( 106/100) gm= 1.06 gm

Mass of solvent=1100-1.06=1098.94 g

Molality= Moles of solution in 1kg of Solvent= {10^(-2)×1000}/1098.94= 9×10^(-3) moles

Answer 2

Given,

density of the solution $\rho = 1.1 g/ml$

$GMW_{Na_2CO_3} = 106$

To find,

Molality(M) of the given solution

Sol:

$10^{-2}$ moles of $(Na_2CO_3) in ----- > 1000$ ml solvent

mass of solution = $voulme_{sol}*density_{sol}= 1000*1.1 =1100 g$

mass of solvent = mass of solution - mass of solute

$m_{solv}=m_{sol}-m_{solute}\\m_{solv}=1100-(M*(GMW_{Na_2CO_3))\\\\m_{solv}= 1100-(10^{-2}*106)\\m_{solv}=1098.94 g$

strength or molality of the solution is

$M = \frac{n_{solute}}{m_{solv}(kg)} \\M =\frac{10^{-2}}{\frac{1098.94}{1000} } \\M = 9.099 *10^{-3} moles$

∴Molality of the solution is $9.099*10^{-3} moles$

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