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Molarity

A toothpaste ...

Question

A toothpaste containing flouride in form of sodium mono fluoroortho phosphate ${N a}_{3} P O_{4} F$ has $0.754$ g sodium in $100$ mL solution. Which of the following are correct for the given observation?

The number of flourine atoms present in

100

mL =

6.62 \times 10^{21}

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The mass of

{N a}_{3} P O_{4} F

100 mL =

2.0

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The concentration of

{N a}_{3} P O_{4} F

0.11

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Mole of

{N a}_{3} P O_{4} F

= mole of F =mole of P

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Solution

The correct options are
A Mole of ${N a}_{3} P O_{4} F$ = mole of F =mole of P
B The mass of ${N a}_{3} P O_{4} F$ 100 mL = $2.0$ g
C The number of flourine atoms present in $100$ mL = $6.62 \times 10^{21}$
D The concentration of ${N a}_{3} P O_{4} F$ = $0.11$ M
Given, 100 ml contains 0.754 g sodium.
So, mass of compound in 100 ml solution is $\frac{183 \times 0.754}{69} = 2.07 g$ .
Molarity of solution is $\frac{2.07 \times 10}{183} = 0.11 M$ .
Mass of fluoride in $100$ ml solution is $\frac{2.07 \times 19}{183} = 0.214 g$ .
Molecules of fluoride $= \frac{0.214}{19} \times N_{A} = 6.8 \times 10^{21}$ .

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A toothpaste containing flouride in form of sodium mono fluoroortho phosphate Na3PO4F has 0.754 g sodium in 100 mL solution. Which of the following are correct for the given observation?

A toothpaste containing flouride in form of sodium mono fluoroortho phosphate ${N a}_{3} P O_{4} F$ has $0.754$ g sodium in $100$ mL solution. Which of the following are correct for the given observation?