seawater contains approximately 3.5% NaCl by mass and has a density of 1.02 g/mL. What volume of seawater contains 1.0 g of sodium?
I would do this.
3.5% NaCl by mass means
3.5g NaCl/100 g solution. The density of th solution is 1.02 g/mL; therefore, the 100 g soln has a volume of mass/density = 100g/1.02 = 98.04 mL.
Next, I would convert 3.5 g NaCl to g Na. That is easily done by
3.5g NaCl x (molar mass Na/molar mass NaCl) = 1.4 g Na. This is an estimate; you can do it more accurately. In reality, then, our solution is 1.4 g Na/98.04 mL.
(1.4 g Na/98.04 mL) x ? mL = 1.0 g Na.
Solve for ?mL.
To find the volume of seawater that contains 1.0 g of sodium, we need to determine the mass of 1.0 g of sodium in seawater. Since seawater contains approximately 3.5% sodium chloride (NaCl) by mass, we can assume that the mass of sodium in seawater is also approximately 3.5% of the total mass of NaCl.
First, let's calculate the mass of NaCl in the given volume of seawater. Since the density of seawater is 1.02 g/mL, this means that 1 mL (or 1 cm³) of seawater has a mass of 1.02 grams.
Next, we need to determine the mass of NaCl in 1 mL of seawater. Since the concentration of NaCl in seawater is 3.5% by mass, we can find the mass of NaCl in 1 mL of seawater by multiplying the density of seawater (1.02 g/mL) by the concentration of NaCl (3.5%):
Mass of NaCl in 1 mL of seawater = (1 mL) * (1.02 g/mL) * (3.5/100)
= 0.0357 g
Therefore, the mass of sodium in 1 mL of seawater would also be approximately 0.0357 g.
Finally, we can determine the volume of seawater that contains 1.0 g of sodium by dividing the given mass (1.0 g) by the mass of sodium in 1 mL of seawater (0.0357 g):
Volume of seawater = 1.0 g / 0.0357 g
≈ 28 mL
Therefore, approximately 28 mL of seawater would contain 1.0 g of sodium.