Question

Can you help me solve these and do them number 4

Answer 1

Potassium nitrate has the formula KNO₃. We can calculate the molar mass from the atomic masses of the element in KNO₃. We have 1 K, 1 N and 3 O, so:

`\begin{gathered} M_K=39.0983g/mol \\ M_N=14.0067g/mol \\ M_O=15.9994g/mol \end{gathered}`

`M_(KNO₃)=1\cdot M_K+1\cdot M_N+3\cdot M_O=\left(1\cdot39.0983+1\cdot14.0067+3\cdot15.9994\right)g/mol=101.1032g/mol`

Since we want a 0.700 M solution with volume of 0.250 L, we can calculate the number of moles of KNO₃ needed:

`\begin{gathered} C=(n_(KNO₃))/(V_(solution)) \\ n_(KNO₃)=C\cdot V_(soution) \\ n_(KNO₃)=0.700mol/L\cdot0.250L=0.175mol \end{gathered}`

Now, we can convert the number of moles to mass using the molar mass, M:

`\begin{gathered} M_(KNO₃)=(m_(KNO₃))/(n_(KNO₃)) \\ m_(KNO₃)=\frac{}{}M_(KNO₃)n_(KNO₃) \\ m_(KNO₃)=101.1032g/mol\cdot0.175mol=17.69306g\approx17.7g \end{gathered}`

So, we would need approximately 17.7g.