# Question #c6c4e

##### 1 Answer

#### Explanation:

We're asked to find

of a solution with density

**MOLALITY:**

Since the **mole fraction** is given as *mole fraction*:

#"mole fraction NaOH" = (0.02color(white)(l)"mol NaOH")/(0.02color(white)(l)"mol NaOH" + 0.98color(white)(l)"mol H"_2"O") = 0.02#

Since there are proportionally **molar mass** of water to find the number of grams (and then kilograms) of water:

#0.98cancel("mol H"_2"O")((18.015cancel("g H"_2"O"))/(1cancel("mol H"_2"O")))((1color(white)(l)"kg H"_2"O")/(10^3cancel("g H"_2"O"))) = color(red)(ul(0.0177color(white)(l)"kg H"_2"O"#

We can now find the **molality**:

#"molality" = "mol solute"/"kg solvent" = (0.02color(white)(l)"mol NaOH")/(color(red)(0.0177color(white)(l)"kg H"_2"O")) = color(blue)(ulbar(|stackrel(" ")(" "1.13color(white)(l)m" ")|)#

**MOLARITY:**

The equation for **molarity** is

#"molarity" = "mol solute"/"L solution"#

We can calculate the molarity from the molality if the solution's **density** is known (given as

What we need to do is convert the **grams** (using its **molar mass**), and add *that* value to the mass of water (equal to **total mass of solution**:

#0.02cancel("mol NaOH")((40.00color(white)(l)"g NaOH")/(1cancel("mol NaOH"))) = color(red)(ul(0.800color(white)(l)"g NaOH"#

The ** total mass of the solution** is thus

#"g solution" = color(red)(0.800color(white)(l)"g NaOH") + 17.7color(white)(l)"g H"_2"O" = color(green)(ul(18.5color(white)(l)"g solution"#

Now we use the **density** of the solution to calculate the number of **liters**:

#color(green)(18.5)cancel(color(green)("g soln"))((1cancel("mL soln"))/(1.1cancel("g soln")))((1color(white)(l)"L soln")/(10^3cancel("mL soln"))) = color(purple)(ul(0.0168color(white)(l)"L solution"#

And the **molarity** is thus

#"molarity" = (0.02color(white)(l)"mol NaOH")/(color(purple)(0.0168color(white)(l)"L solution")) = color(blue)(ulbar(|stackrel(" ")(" "1.19color(white)(l)M" ")|)#