Respuesta :
m(NaCl)=1.0 g
w=0.035
p=1.02 g/mL
v-?
m(NaCl)=mw
m=vp
m(NaCl)=vpw
v=m(NaCl)/(pw)
v=1.0/(1.02·0.035)=28.0 mL
w=0.035
p=1.02 g/mL
v-?
m(NaCl)=mw
m=vp
m(NaCl)=vpw
v=m(NaCl)/(pw)
v=1.0/(1.02·0.035)=28.0 mL
Answer:
28 mL
Explanation:
We want to know what volume of seawater contains 1.0 g of sodium.
We know that the total amount of water mass always 3.5% corresponds to the mass of sodium.
So if the total mass of sodium is 1.0g this is equivalent to 3.5% of 100% of the mass
Calculate 100% of the mass with a simple rule of three
[tex]3.5\% \longrightarrow 1.0g\\100\% \longrightarrow x\\x=\frac{(100)(1)}{3.5}\\ x=28.57 g[/tex]
The total mass of seawater containing 1 g of sodium is 28.57 g
Now we calculate the volume that corresponds to this mass with the density
[tex]d=\frac{m}{v} \\v=\frac{m}{d}\\[/tex]
replace the values in the formula
[tex]v=\frac{28.57}{1.02}\\ v= 28 mL[/tex]
