QuestionCalculate the density of a liquid in \$Lbs/in^{3}\$ given its mass and volume. Also, find the volume of 150 lbs of MgCl\(_2 \cdot 6\)H\(_2\)O.

Studdy Solution

STEP 1

Assumptions1. The volume of the unknown liquid added is $17.8 \, mL$ . The weight of the empty flask is $354.7 \, g$
3. The total mass of the flask and the liquid is $462.4 \, g$
4. We need to find the density of the liquid in $\mathrm{bs} / \mathrm{in}^{3}$

STEP 2

First, we need to find the mass of the liquid. We can do this by subtracting the weight of the empty flask from the total mass.
$Mass_{liquid} = Total\, mass - Mass_{flask}$

STEP 3

Now, plug in the given values for the total mass and the mass of the flask to calculate the mass of the liquid.
$Mass_{liquid} =462.\, g -354.7\, g$

STEP 4

Calculate the mass of the liquid.
$Mass_{liquid} =462.4\, g -354.7\, g =107.7\, g$

STEP 5

Now that we have the mass of the liquid, we can find the density of the liquid. The density is defined as the mass divided by the volume.
$Density_{liquid} = \frac{Mass_{liquid}}{Volume_{liquid}}$

STEP 6

Plug in the values for the mass and the volume of the liquid to calculate the density.
$Density_{liquid} = \frac{107.\, g}{17.8\, mL}$

STEP 7

Calculate the density of the liquid in $g/mL$ .
$Density_{liquid} = \frac{107.7\, g}{17.\, mL} =6.05\, g/mL$

STEP 8

Now, we need to convert the density from $g/mL$ to $\mathrm{bs} / \mathrm{in}^{3}$ . We know that $1\, g =0.00220462\, lbs$ and $1\, mL =0.0000610237\, in^{3}$ .
$Density_{liquid} =6.05\, g/mL \times0.00220462\, lbs/g \times1/0.0000610237\, in^{3}/m$

STEP 9

Calculate the density of the liquid in $\mathrm{bs} / \mathrm{in}^{3}$ .
$Density_{liquid} =6.05\, g/mL \times.00220462\, lbs/g \times/.0000610237\, in^{3}/m =.000221\, \mathrm{bs} / \mathrm{in}^{3}$ The density of the liquid is $.000221\, \mathrm{bs} / \mathrm{in}^{3}$ .

STEP 10

Assumptions for the second problem. The density of Magnesium chloride hexahydrate $(\mathrm{MgCl}_{2}{ }^{}6 \mathrm{H}_{2} \mathrm{})$ is $.69 \, g / cm^{3}$
2. The mass of $\mathrm{MgCl}_{2}{ }^{}6 \mathrm{H}_{2} \mathrm{}$ is $.5 \times10^{2}$ pounds3. We need to find the volume in liters

STEP 11

First, we need to convert the mass from pounds to grams. We know that $\, lb =453.592\, g$ .
$Mass_{MgCl_{}.6H_{}} =.5 \times10^{}\, lbs \times453.592\, g/lb$

STEP 12

Calculate the mass of $\mathrm{MgCl}_{2}{ }^{*}6 \mathrm{H}_{2} \mathrm{}$ in grams.
$Mass_{MgCl_{2}.6H_{2}} =.5 \times10^{2}\, lbs \times453.592\, g/lb =6.8 \times10^{4}\, g$

STEP 13

Now that we have the mass of $\mathrm{MgCl}_{2}{ }^{*}6 \mathrm{H}_{2} \mathrm{}$ , we can find the volume. The volume is defined as the mass divided by the density.
$Volume_{MgCl_{2}.6H_{2}} = \frac{Mass_{MgCl_{2}.6H_{2}}}{Density_{MgCl_{2}.6H_{2}}}$

STEP 14

Plug in the values for the mass and the density of $\mathrm{MgCl}_{2}{ }^{*}6 \mathrm{H}_{2} \mathrm{}$ to calculate the volume.
$Volume_{MgCl_{2}.6H_{2}} = \frac{6.8 \times10^{4}\, g}{.69\, g/cm^{3}}$

STEP 15

Calculate the volume of $\mathrm{MgCl}_{2}{ }^{*} \mathrm{H}_{2} \mathrm{}$ in $cm^{3}$ .
$Volume_{MgCl_{2}.H_{2}} = \frac{.8 \times10^{4}\, g}{.69\, g/cm^{3}} =4.02 \times10^{4}\, cm^{3}$

STEP 16

Now, we need to convert the volume from $cm^{3}$ to liters. We know that $\, cm^{3} =0.001\, L$ .
$Volume_{MgCl_{2}.6H_{2}} =4.02 \times10^{4}\, cm^{3} \times0.001\, L/cm^{3}$

STEP 17

Calculate the volume of $\mathrm{MgCl}_{2}{ }^{}6 \mathrm{H}_{2} \mathrm{}$ in liters.
$Volume_{MgCl_{2}.6H_{2}} =4.02 \times10^{4}\, cm^{3} \times0.001\, L/cm^{3} =40.2\, L$ The volume of $.5 \times10^{2}$ pounds of $\mathrm{MgCl}_{2}{ }^{}6 \mathrm{H}_{2} \mathrm{}$ is $40.2\, L$ .

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