Question

the density of 3M NA2CO3 IS is 1.2g ml. the percentage by weight and ppm of NA2CO3 will be?​

Answer 1

Answer:

3 moles of Na

2

S

2

O

3

=3×158=474g

Volume= 1000

ρ=1.56g/ml, weight= 1560g

(i) Weight %=

1560

474

=30.38

(ii) Mole fraction, moles of solvent=

18

W

=

18

1560−474

=60.83 moles

Mole fraction of Na

2

S

2

O

3

=

3+60.83

3

=0.046

(iii) Molality=

1.086

3

=2.76

Molality of Na

+

=2×2.76=5.52m

Molality of S

2

O

3

2−

=2.76m

Answer 2

Percentage By Weight Of Na₂co₃ Is 26.5% And Parts Per Million Is 2.65×10⁵ Ppm

GIVEN :

Moles of Na₂CO₃ = 3 M

Density = 1.2 g/mL

TO FIND

Percentage by weight

parts per million of Na₂CO₃

SOLUTION

We can simply solve the problem as under

Let the volume of solution be, V

Density of solution = Mass /volume = 1.2

Molarity = moles of solute / volume of solution

Number of moles of Na₂CO₃ = 3 × V × 10⁻³

Molar mass of Na₂CO₃ = 3(23)+12+(16)3

Molar mass = 106 grams

Total Weight of Na₂CO₃ = 3 × V× 10⁻³ × 106

Total weight of Solution = 1.2V gm

Mass % of Na₂CO₃ = (weight of solute /total weight of solution ) × 100

Mass % of Na₂CO₃ =

$\frac{3 \times V \times {10}^{ - 3 \: \times 106} }{1.2V} \times 100$

% weight of Na₂CO₃ = 26.5%

% weight of Na₂CO₃(ppm) =

$\frac{3 \times V \times {10}^{ - 3 \: \times 106} }{1.2V} \times {10}^{6}$

= 2.65 × 10⁵ ppm.

Hence, percentage by weight of Na₂CO₃ is 26.5% and parts per million is 2.65×10⁵ ppm

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