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Some Basic Concepts of Chemistry question

2021 · 26 Aug · Shift 2 · Q18
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Some Basic Concepts of Chemistry question

2021 · 26 Aug · Shift 2 · Q18

JEE MainChemistrySome Basic Concepts of ChemistryNumerical+4 / −1
100 mL of Na3PO4Na_3PO_4Na3​PO4​ solution contains 3.45 g of sodium. The molarity of the solution is ‾×\underline{\hspace{2cm}}\times​× 10 −-− 2 mol L −-− 1. (Nearest integer) [Atomic Masses - Na : 23.0 u, O : 16.0 u, P : 31.0 u]
Numerical answer
View written solutionFree

Correct answer: 50

  1. Given data
  • Volume of solution =100 mL=0.1 L= 100\,\text{mL} = 0.1\,\text{L}=100mL=0.1L
  • Mass of sodium present =3.45 g= 3.45\,\text{g}=3.45g
  • Solute is Na3PO4Na_3PO_4Na3​PO4​
  1. Find moles of sodium atoms

Atomic mass of sodium =23 g mol−1= 23\,\text{g mol}^{-1}=23g mol−1

n(Na)=3.4523=0.15 moln(\text{Na}) = \frac{3.45}{23} = 0.15\,\text{mol}n(Na)=233.45​=0.15mol

  1. Relate moles of sodium to moles of Na3PO4Na_3PO_4Na3​PO4​

Each mole of Na3PO4Na_3PO_4Na3​PO4​ contains 333 moles of Na.

So,

n(Na3PO4)=0.153=0.05 moln(Na_3PO_4) = \frac{0.15}{3} = 0.05\,\text{mol}n(Na3​PO4​)=30.15​=0.05mol

  1. Find molarity of the solution

This amount is present in 0.1 L0.1\,\text{L}0.1L.

M=0.050.1=0.5 mol L−1M = \frac{0.05}{0.1} = 0.5\,\text{mol L}^{-1}M=0.10.05​=0.5mol L−1

  1. Match with the asked format

The molarity is written as:

‾×10−2 mol L−1\underline{\hspace{2cm}} \times 10^{-2}\,\text{mol L}^{-1}​×10−2mol L−1

Let the blank be xxx. Then,

x×10−2=0.5x \times 10^{-2} = 0.5x×10−2=0.5

x=0.5×102=50x = 0.5 \times 10^2 = 50x=0.5×102=50

  1. Final answer

50\boxed{50}50​

The nearest integer is 505050.

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