A $5.95 \cdot g$ $5.95g$ mass of potassium bromide were dissolved in $400 \cdot c m^{3}$ $400cm^3$ water ... what was the concentration of the solution in $m o l \cdot L^{- 1}$ $mol*L^-1$ ?

Question

A $5.95 \cdot g$ $5.95g$ mass of potassium bromide were dissolved in $400 \cdot c m^{3}$ $400cm^3$ water ... what was the concentration of the solution in $m o l \cdot L^{- 1}$ $mol*L^-1$ ?

1 Answer

Impact of this question

4184 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-14T07:40:07

Approx. $0.1 \cdot m o l \cdot L^{- 1}$ $0.1*mol*L^-1$ ......................

Explanation:

We use the quotient..... $Molarity \equiv \frac{Moles of solute}{Volume of solution}$ $"Molarity"-="Moles of solute"/"Volume of solution"$ , and we know that $1 \cdot c m^{3} \equiv 1 \cdot m L \equiv 10^{- 3} \cdot L$ $1*cm^3-=1*mL-=10^-3*L$ ......

And thus $Molarity = \frac{\frac{5.95 \cdot g}{119.0 \cdot g \cdot m o l^{- 1}}}{400 \cdot c m^{3} \times 10^{- 3} \cdot L \cdot c m^{- 3}}$ $"Molarity"=((5.95*g)/(119.0*g*mol^-1))/(400*cm^3xx10^-3*L*cm^-3)$

$= 0.125 \cdot m o l \cdot L^{- 1}$ $=0.125*mol*L^-1$ with respect to $K B r$ $KBr$ . What are the concentrations with respect to $K^{+}$ $K^+$ and $B r^{-}$ $Br^-$ ? Are the units I use correct?

Science

Math

Humanities

... and beyond

A $5.95 \cdot g$ $5.95g$ mass of potassium bromide were dissolved in $400 \cdot c m^{3}$ $400cm^3$ water ... what was the concentration of the solution in $m o l \cdot L^{- 1}$ $mol*L^-1$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

A 5.95⋅g5.95*g mass of potassium bromide were dissolved in 400⋅cm3400*cm^3 water ... what was the concentration of the solution in mol⋅L−1mol*L^-1?

1 Answer

Explanation:

Related questions

Impact of this question

A $5.95 \cdot g$ $5.95g$ mass of potassium bromide were dissolved in $400 \cdot c m^{3}$ $400cm^3$ water ... what was the concentration of the solution in $m o l \cdot L^{- 1}$ $mol*L^-1$ ?