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k0ka [10]
3 years ago
5

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>

Molarity of lead (II) nitrate solution = 2.70 M

Volume of solution = 33.0 mL = 0.033 L   (Conversion factor: 1 L = 1000 mL)

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

4 0
3 years ago
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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When equal volumes of 0.5 M HCl and 0.5 M Ca(OH)2 are mixed, the resulting solution is
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The concentration of mixed solution = 0.5 M

<h3> Further explanation </h3>

Given

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2V.Mm = V(M₁+M₂)

2V.Mm = V(0.5+0.5)

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