What volume of $0.315 \cdot m o l \cdot L^{- 1}$ $0.315molL^-1$ $N a O H$ $NaOH$ is required to deliver $6.22 \cdot g$ $6.22*g$ of $N a O H$ $NaOH$ ?

Question

What volume of $0.315 \cdot m o l \cdot L^{- 1}$ $0.315molL^-1$ $N a O H$ $NaOH$ is required to deliver $6.22 \cdot g$ $6.22*g$ of $N a O H$ $NaOH$ ?

1 Answer

Impact of this question

4564 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-12T08:14:52

Approx. half a litre of the given solution is required.........

Explanation:

We require $\frac{6.22 \cdot g}{40.00 \cdot g \cdot m o l^{- 1}} = 0.0156 \cdot m o l$ $(6.22*g)/(40.00*g*mol^-1)=0.0156*mol$ WITH RESPECT TO $N a O H$ $NaOH$ .

Now, by definition, $Concentration = \frac{Moles}{Volume}$ $"Concentration"="Moles"/"Volume"$ ......

And thus $Volume = \frac{Moles}{Concentration} = \frac{0.0156 \cdot m o l}{0.315 \cdot m o l \cdot L^{- 1}}$ $"Volume"="Moles"/"Concentration"=(0.0156*mol)/(0.315*mol*L^-1)$

$= 0.494 \cdot L$ $=0.494*L$ .

Science

Math

Humanities

... and beyond

What volume of $0.315 \cdot m o l \cdot L^{- 1}$ $0.315molL^-1$ $N a O H$ $NaOH$ is required to deliver $6.22 \cdot g$ $6.22*g$ of $N a O H$ $NaOH$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

What volume of 0.315*mol*L^-10.315⋅mol⋅L−10.315*mol*L^-1 NaOHNaOHNaOH is required to deliver 6.22*g6.22⋅g6.22*g of NaOHNaOHNaOH?

1 Answer

Explanation:

Related questions

Impact of this question

What volume of $0.315 \cdot m o l \cdot L^{- 1}$ $0.315molL^-1$ $N a O H$ $NaOH$ is required to deliver $6.22 \cdot g$ $6.22*g$ of $N a O H$ $NaOH$ ?