Answer to Question #162554 in General Chemistry for Sage Topaz

Question

Calculate the molality of the a. 0.050 g CO2 in 652 g of H2O, b. 56 g NH3 in 200 g of H2O, c. 3.21 g C6H12 in 2.3 g of benzene (C6H6).

Answer 1

a) Moles of $CO_2=$ $0.050CO_2\times$ $1 molCO_2\over 44.01gCO_2$ $=0.0011molCO_2$

$1g=0.001L$

Liter of solution=$0.652L$

Molarity$=$ $Moles of solute\over liter of solution$

$=$ $0.0011mol \over 0.652L$ $0.00174mol/L$

b) Moles of $NH_3=56gNH_3\times$ $1mol NH_3\over 17.03gNH_3$ $=3.289mol NH_3$

Liter of solution$=0.2L$

Molarity$=Moles of solute\over liter of solution$

$=3.289mol\over 0.2L$ $=16.44mol/L$

c) Moles of $C_6H_{12}=$ $3.21gC_6H_{12}\times$ $1molC_6H_{12}\over 84.16gC_6H_{12}$ $=0.0381molC_6H_{12}$

Liter of solution$=0.0023L$

Molarity$=$ $Moles of dilute\over liter of solution$

$=$ $0.0381mol\over 0.0023L$ $16.56 mol/L$