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3 years ago

Calculate the amount of water required to prepare 500g of 2.5% solution of sugar.

Chemistry

1 answer:

Snowcat [4.5K]3 years ago

8 0

(i) We start by calculating the mass of sugar in the solution:

mass of sugar = concentration × solution mass

mass of sugar = 2.5/100 × 500 = 12.5 g

Then now we can calculate the amount of water:

solution mass = mass of sugar + mass of water

mass of water = solution mass - mass of sugar

mass of water = 500 - 12.5 = 487.5 g

(ii) We use the following reasoning:

If 500 g solution contains 12.5 g sugar

Then X g solution contains 75 g sugar

X=(500×75)/12.5 = 3000 g solution

Now to get the amount of solution in liters we use density (we assume that is equal to 1):

Density = mass / volume

Volume = mass / density

Volume = 3000 / 1 = 3000 liters of sugar solution

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A student mixes 33.0 mL of 2.70 M Pb ( NO 3 ) 2 ( aq ) with 20.0 mL of 0.00157 M NaI ( aq ) . How many moles of PbI 2 ( s ) prec

GalinKa [24]

<u>Answer:</u> The moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

<u>For lead (II) nitrate:</u>

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Putting values in equation 1, we get:

$2.70M=\frac{\text{Moles of lead (II) nitrate}}{0.033L}\\\\\text{Moles of lead (II) nitrate}=(2.70mol/L\times 0.0330L)=0.0891mol$

<u>For NaI:</u>

Molarity of NaI solution = 0.00157 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

$0.00157M=\frac{\text{Moles of NaI}}{0.020L}\\\\\text{Moles of NaI}=(0.00157mol/L\times 0.0200L)=3.14\times 10^{-5}mol$

For the given chemical reaction:

$Pb(NO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaNO_3(aq.)$

By Stoichiometry of the reaction:

2 moles of NaI reacts with 1 mole of lead (II) nitrate

So, $3.14\times 10^{-5}$ moles of NaI will react with = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}mol$ of lead (II) nitrate

As, given amount of lead (II) nitrate is more than the required amount. So, it is considered as an excess reagent.

Thus, NaI is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, $3.14\times 10^{-5}$ moles of NaI will produce = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}moles$ of lead (II) iodide

Hence, the moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

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PLZ HELP I WILL GIVE U BRAINLEST PLZ HELP!!!!!!!!!!

sergeinik [125]

My answer is A. I'm probably wrong. in bad in this subject

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