How many grams are needed to prepare 750. mL of a 0.35 M solution of NaOH?

Chemistry

1 answer:

nikdorinn [45]3 years ago

3 0

Answer:

10.5grams

Explanation:

Molarity = number of moles (n)/ volume (V)

According to this question;

Volume = 750 mL = 750/1000 = 0.75L

Molarity = 0.35M

number of moles (n) = molarity × volume

n = 0.35 × 0.75

n = 0.2625mol

Using mole = mass/molar mass

Molar Mass of NaOH = 23 + 16 + 1

= 40g/mol

mole = mass/molar mass

0.2625 = mass/40

mass = 10.5grams

10.5 grams are needed to prepare 0.75L of a 0.35 M solution of NaOH.

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When 125 mL of 0.150 M Pb(NO3)2 is mixed with 145 mL of 0.200 M KBr, 4.92 g of PbBr2 is collected. Calculate the percent yield.

Semenov [28]

Answer:

Y = 92.5 %

Explanation:

Hello there!

In this case, since the reaction between lead (II) nitrate and potassium bromide is:

$Pb(NO_3)_2+2KBr\rightarrow PbBr_2+2KNO_3$

Exhibits a 1:2 mole ratio of the former to the later, we can calculate the moles of lead (II) bromide product to figure out the limiting reactant:

$0.125L*0.150\frac{molPb(NO_3)_2}{L} *\frac{1molPbBr_2}{1molPb(NO_3)_2} =0.01875molPbBr_2\\\\0.145L*0.200\frac{molKBr}{L} *\frac{1molPbBr_2}{2molKBr} =0.0145molPbBr_2$

Thus, the limiting reactant is the KBr as it yields the fewest moles of PbBr2 product. Afterwards, we calculate the mass of product by using its molar mass:

$0.0145molPbBr_2*\frac{367.01gPbBr_2}{1molPbBr_2} =5.32gPbBr_2$