91Ó°ÊÓ

We are given a toothpaste tube of $119g$containing $0.240\%$by mass of $\mathrm{NaF}$.

We need to find moles of$\mathrm{NaF}$in toothpaste.

Amount of toothpaste=$119g$

Composition of $\mathrm{NaF}$in toothpaste= localid="1653457114580" $0.240\%=\frac{0.240×119g}{100}=0.2856g$

Number of moles =$\frac{Weightofsubs\tan ce}{Molecularmassofsubs\tan ce}$

So, number of moles in localid="1652505003670" $\mathrm{NaF}$= localid="1653461551129" $\frac{0.2856\cancel{g}}{42\cancel{g}}=0.068$

Hence, localid="1652504997441" $0.0068$moles of localid="1652505007178" $\mathrm{NaF}$is present in toothpaste.

We are given a toothpaste tube of $119g$containing $0.240\%$by mass of $\mathrm{NaF}$.

We need to find how many fluoride ions are in toothpaste.

• Moles of $\mathrm{NaF}=0.0068$(Calculated in part (a) step $2$)
• Number of ions= Number of moles* Avogadro’s number
• Number of ${\mathrm{F}}^{-}$ions in toothpaste = localid="1652505045894" $0.0068×(6.022×{10}^{23})=4.09×{10}^{21}$

Hence, in one toothpaste tube localid="1652505055726" $4.09×{10}^{21}$fluoride ions are present.

We are given a toothpaste tube of$119g$containing $0.240\%$by mass of $\mathrm{NaF}$.

We need to find how many grams of ${\mathrm{Na}}^{+}$ions are in toothpaste.

• Composition of $\mathrm{NaF}=\frac{0.240×1.50g}{100}=0.036\mathrm{g}$

• Moles of $\mathrm{NaF}$in $1.50g$of toothpaste = $\frac{0.036\cancel{g}}{42\cancel{g}}=0.00085$
• Number of ions of $\mathrm{Na}$=$0.00085×6.022×{10}^{23}=5.1×{10}^{20}$

Hence, Grams of sodium ion= localid="1653456969713" $5.1×{10}^{20}×23g=1.17×{10}^{22}g$

We are given that in toothpaste tube of $119g$, $5\%$by mass of ${\mathrm{KNO}}_3$is present.

We need to find formula unit of ${\mathrm{KNO}}_3$present in toothpaste.

• Composition by mass of ${\mathrm{KNO}}_3$= $\frac{5×119}{100}=5.95g$
• Moles of localid="1653457012769" ${\mathrm{KNO}}_3=\frac{5.95\cancel{g}}{101\cancel{g}}=0.058$
• Number of formula units = Number of moles * Avogadro’s number

$$\begin{gathered} =0.058×6.022×{10}^{23} \\ =0.034×{10}^{23} \\ =3.4×{10}^{21} \end{gathered}$$

Hence, number of formula units of${\mathrm{KNO}}_3$is$3.4×{10}^{21}$.