What is the concentration of a $7.8 \times 10^{- 9} \cdot m o l \cdot L^{- 1}$ $7.8xx10^-9molL^-1$ solution of chloroacetic acid in $parts per billion$ $"parts per billion"$ , i.e. $μ g \cdot L^{- 1}$ $mug*L^-1$ ?

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What is the concentration of a $7.8 \times 10^{- 9} \cdot m o l \cdot L^{- 1}$ $7.8xx10^-9molL^-1$ solution of chloroacetic acid in $parts per billion$ $"parts per billion"$ , i.e. $μ g \cdot L^{- 1}$ $mug*L^-1$ ?

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Answer 1 · 2017-06-28T15:37:28

Approx....... $0.74 \cdot μ g \cdot L^{- 1}$ $0.74*mug*L^-1$ with respect to $chloroacetic acid......$ $"chloroacetic acid......"$

Explanation:

$7.8$ $7.8$ $nanomoles per litre$ $"nanomoles per litre"$ $=$ $=$ $7.8 \times 10^{- 9} \cdot m o l \cdot L^{- 1}$ $7.8xx10^-9*mol*L^-1$ .

$= 94.9 \cdot g \cdot m o l^{- 1} \times 7.8 \times 10^{- 9} \cdot m o l \cdot L^{- 1}$ $=94.9*g*mol^-1xx7.8xx10^-9*mol*L^-1$

$= 7.40 \times 10^{- 7} \cdot g \cdot L^{- 1}$ $=7.40xx10^-7*g*L^-1$

$= 7.40 \times 10^{- 7} \cdot g \cdot L^{- 1} \times 10^{6} \cdot μ g \cdot g^{- 1}$ $=7.40xx10^-7*g*L^-1xx10^6*mug*g^-1$

$=0.74xxmug*L^-1.......$

Answer 2 · 2017-06-28T15:53:27

$0.74$ $mu"g L"^(-1)$

Explanation:

The idea here is that you need to convert the number of nanomoles of chloroacetic acid, or monochloroacetic acid (MCA), to micrograms by using the molar mass of the acid.

The problem provides you with the chemical formula of MCA, so grab a Periodic Table and calculate its molar mass.

You know that $1$ mole of MCA contains

two moles of carbon, $2 xx "C"$

two moles of oxygen, $2 xx "O"$

three moles of hydrogen, $3 xx "H"$

one mole of chlorine, $1 xx "Cl"$

To find the molar mass of the acid, simply add the molar masses of all the atoms that make up the compound.

You should end up with

$2 * "12.011 g mol"^(-1) + 2 * "15.994 g mol"^(-1) + 3 * "1.00794 g mol"^(-1) + 1 * "35.453 g mol"^(-1)$

$= "94.4976 g mol"^(-1)$

So, you know that $1$ mole of MCA has a mass of $"94.4976 g"$ and that

$"1 mole" = 10^9$ $"nanomoles"$

This means that $7.8$ nanomoles will have a mass of

$7.8 color(red)(cancel(color(black)("nanomoles MCA"))) * (1color(red)(cancel(color(black)("mole"))))/(10^9color(red)(cancel(color(black)("nanomoles")))) * "94.4976 g"/(1color(red)(cancel(color(black)("mole MCA"))))$

$= 737.08 * 10^(-9)$ $"g"$

So, you know that

$"7.8 nanomoles L"^(-1) = 737.08 * 10^(-9)$ $"g L"^(-1)$

and that

$"1 g" = 10^6$ $mu"g"$

so you can say that the concentration of the acid expressed in micrograms per liter will be equal to

$737.08 * 10^(-9) color(red)(cancel(color(black)("g")))"L"^(-1) * (10^6color(white)(.)mu"g")/(1color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)(0.74color(white)(.)mu"g L"^(-1))))$

The answer is rounded to two sig figs, the number of sig figs you have for the given concentration.

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What is the concentration of a $7.8 \times 10^{- 9} \cdot m o l \cdot L^{- 1}$ $7.8xx10^-9molL^-1$ solution of chloroacetic acid in $parts per billion$ $"parts per billion"$ , i.e. $μ g \cdot L^{- 1}$ $mug*L^-1$ ?

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What is the concentration of a $7.8 \times 10^{- 9} \cdot m o l \cdot L^{- 1}$ $7.8xx10^-9molL^-1$ solution of chloroacetic acid in $parts per billion$ $"parts per billion"$ , i.e. $μ g \cdot L^{- 1}$ $mug*L^-1$ ?