Chapter 14.2, Problem 16PP

Interpretation Introduction

Interpretation:

Molarity of an aqueous solution of $40gm$ glucose in $1.5L$ solution is to be calculated

Concept introduction:

The concentrations of solutions are expressed in different ways, all of them involved in the quantity of solute and the quantity of solution or solvent. Among them, molarity is widely used to express the concentration of solution. In expression, $\mathrm{Molarity}ofsolution=\frac{Numberofmolesofsolute}{Volumeofsolution(inlitre)}$

Again $\mathrm{Number}ofmolesofsolute=\frac{Massofsolute}{molecularweightofsolute}$

Answer to Problem 16PP

Explanation:

In expression, $\mathrm{Molarity}ofsolution=\frac{Numberofmolesofsolute}{Volumeofsolution(inlitre)}$

Again $\mathrm{Number}ofmolesofsolute=\frac{Massofsolute}{molecularweightofsolute}$

$=\frac{Massofglucose}{Molecularweightofglucose}$

(Data given: Mass of glucose = $40gm$ )

$=\frac{40}{180.16}$ (Molecular weight of glucose = $180.16gm$ )

= $0.222$

So, molarity of 0.222 mole glucose solution $=\frac{0.222}{1.5}$

= $0.148M$

Molarity of an aqueous solution of $40gm$ glucose in $1.5L$ solution is $0.148M$

Explanation of Solution

In expression, $\mathrm{Molarity}ofsolution=\frac{Numberofmolesofsolute}{Volumeofsolution(inlitre)}$

Again $\mathrm{Number}ofmolesofsolute=\frac{Massofsolute}{molecularweightofsolute}$

$=\frac{Massofglucose}{Molecularweightofglucose}$

(Data given: Mass of glucose = $40gm$ )

$=\frac{40}{180.16}$ (Molecular weight of glucose = $180.16gm$ )

= $0.222$

So, molarity of 0.222 mole glucose solution $=\frac{0.222}{1.5}$

= $0.148M$

Conclusion

Molarity of an aqueous solution of $40gm$ glucose in $1.5L$ solution is $0.148M$