## Chemistry 12th Edition

We know that: (a) $Mr\space of\space CH_3OH=32$ $Moles=\frac{6.57}{32}=0.205 \space moles$ $Molarity=\frac{0.205}{0.150}=1.37\frac{mol}{L}$ (b) $Mr\space of\space CaCl_2=111$ $Moles=\frac{10.4}{111}=0.0937 \space moles$ $Molarity=\frac{0.093moles}{0.22L}=0.42\frac{mol}{L}$ (c) $Mr\space of\space C_{10}H_8=128$ $Moles=\frac{7.82}{128}=0.061 \space moles$ $Molarity=\frac{0.061moles}{0.0852L}=0.71\frac{mol}{L}$