Question

Question #54da7

Answer 1

`=1.4283M`

Explanation:

1. Assume the following data for the density of water and nitric acid.
   `HNO_3=1.42" g/ml"` `("concentrated"`)
   `H_2O=1.00" g/ml"` `("at " 4^oC`)
2. Since the provided data are in grams, convert it to volume. Observe uniformity of units; i.e.,
   `V_(HNO_3)=(630cancel(g))/((1.42cancel(g))/(ml))=443.6620mL~~443.7mL`
   `V_(H_2O)=(6500cancel(g))/((1.00cancel(g))/(ml))=6500mL`
3. Find the volume of the solution. Note that volume solution is the summation of the volume of the solute and the solvent as shown below.
   `"Volume"_("solution")="volume solute"+"volume solvent"`

`"Volume"_("solution")=443.7mL+6500mL`

`"Volume"_("solution")=6943.7cancel(mL)xx(1L)/(1000cancel(mL))=6.9437L~~7.0000L`
4. Find the mole value of nitric acid `(etaHNO_3)` by taking the quotient of its `"mass"=630g` against its `"molar mass"=63.012" g/mol"`; so that,
`eta=(mHNO_3)/(MmHNO_3)`
`eta=(630g)/((63.012g)/(mol))`
`eta=9.998mol`
5. From the formula `"Molarity"(M)=("number of moles"(eta))/("Li solution"(V)`, find the concentration of the solution.
`M=eta/V`
`M=(9.998mol)/(7.0000L)`
`M=1.4283" mol/L"~~1.4283M`