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:
[tex]\begin{gathered} M_K=39.0983g/mol \\ M_N=14.0067g/mol \\ M_O=15.9994g/mol \end{gathered}[/tex]
[tex]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[/tex]
Since we want a 0.700 M solution with volume of 0.250 L, we can calculate the number of moles of KNO₃ needed:
[tex]\begin{gathered} C=\frac{n_{KNO₃}}{V_{solution}} \\ n_{KNO₃}=C\cdot V_{soution} \\ n_{KNO₃}=0.700mol/L\cdot0.250L=0.175mol \end{gathered}[/tex]
Now, we can convert the number of moles to mass using the molar mass, M:
[tex]\begin{gathered} M_{KNO₃}=\frac{m_{KNO₃}}{n_{KNO₃}} \\ m_{KNO₃}=\frac{}{}M_{KNO₃}n_{KNO₃} \\ m_{KNO₃}=101.1032g/mol\cdot0.175mol=17.69306g\approx17.7g \end{gathered}[/tex]
So, we would need approximately 17.7g.
