Question

250cm^3 solution of sodium sulphate contains 3.01x10^22 sodium ions what is the molality of the solutions

Answer 1

tá súil tá sé ar stailc a gcuid eile den scéal úd

Answer 2

The molarity of sodium sulfate solution is 0.1 M

Explanation:

According to mole concept:

$6.022\times 10^{23}$ number of particles are present in 1 mole of a compound

1 mole of $Na_2SO_4$ contains 2 moles of sodium ions and 1 mole of sulfate ions

Number of sodium ions in 1 mole of $Na_2SO_4=(2\times 6.022\times 10^{23})=12.044\times 10^{23}$

We are given:

Number of sodium ions = $3.01\times 10^{22}$

Applying unitary method:

$12.044\times 10^{23}$ number of sodium ions are present in 1 mole of sodium sulfate

So, $3.01\times 10^{22}$ number of sodium ions will be present in = $\frac{1}{12.044\times 10^{23}}\times 3.01\times 10^{22}=0.025mol$ of sodium sulfate

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

Moles of sodium sulfate = 0.025 moles

Volume of solution = $250cm^3=250mL$      (Conversion factor:  $1cm^3=1mL$ )

Putting values in above equation, we get:

$\text{Molarity of sodium sulfate}=\frac{0.025\times 1000}{250}\\\\\text{Molarity of solution}=0.1M$

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