Question #ac840

2 Answers
Aug 17, 2017

#"m"# #= 1.98# #"kg"#

Explanation:

Yes you can.

The formula for density is #rho = frac("m")("V")#; where #rho# is the density, #"m"# is the mass, and #"V"# is the volume.

Let's substitute the relevant values into the equation:

#Rightarrow 0.793# #"kg L"^(- 1) = frac("m")(2.5 " L")#

#Rightarrow 0.793# #"kg L"^(- 1) times 2.5# #"L" = frac("m")(2.5 " L") times 2.5# #"L"#

#Rightarrow 1.9825# #"kg"# #=# #"m"#

#therefore# #"m" = 1.98# #"kg"#

Therefore, the mass of solute is #1.98# #"kg"#.

Aug 17, 2017

The mass of solute is #"2.0 kg"#.

Refer to the explanation for the process.

Explanation:

Another way of expressing the density formula is:

#D=M/V#, where D is density, M is mass, and V is volume.

(It doesn't hurt to know that the Greek lowercase letter rho, #rho#, is a symbol for density. The Latin D is also acceptable.)

The question gives volume and density, and asks for the mass, which is unknown.

Solution

Rearrange the density formula to isolate #M#. Plug in the known values and solve.

#M=DxxV#

#M="0.793 kg"/color(red)cancel(color(black)("L"))xx2.5color(red)cancel(color(black)("L"))="2.0 kg"# (rounded to two significant figures due to 2.5 L)