1.00 mL of a 3.05 × 10–4 M solution of oleic acid is diluted with 9.00 mL of petroleum ether, forming solution A. Then 2.00 mL of solution A is diluted with 8.00 mL of petroleum ether, forming solution B. How many grams of oleic acid are 5.00 mL of solution B? (Molar mass for oleic acid = 282 g/mol)

Concentration of solution B is 7.10\times 10^{-6}M7.10×10

−6

M

Explanation:

Solution A: Total volume of solution A = (9.00+1.00) mL = 10.00 mL

According to law of dilution, C_{1}V_{1}=C_{2}V_{2}C

1

V

1

=C

2

V

2

where C_{1}C

1

and C_{2}C

2

are initial and final concentration respectively. V_{1}V

1

and V_{2}V

2

are initial and final volume respectively.

Here C_{1}C

1

= 3.55\times 10^{-4}M3.55×10

−4

M , V_{1}V

1

= 1.00 mL, V_{2}V

2

= 10.00 mL

So, C_{2}=\frac{C_{1}V_{1}}{V_{2}}C

2

=

V

2

C

1

V

1

= \frac{3.55\times 10^{-4}M\times 1.00mL}{10.00mL}=3.55\times 10^{-5}M

10.00mL

3.55×10

−4

M×1.00mL

=3.55×10

−5

M

So, concentration of solution A = 3.55\times 10^{-5}M3.55×10

−5

M

Solution B: Total volume of solution B = (2.00+8.00) mL = 10.00 mL

Similarly as above, C_{2}=\frac{C_{1}V_{1}}{V_{2}}C

2

=

V

2

C

1

V

1

= \frac{3.55\times 10^{-5}M\times 2.00mL}{10.00mL}=7.10\times 10^{-6}M

10.00mL

3.55×10

−5

M×2.00mL

=7.10×10

−6

M

So, concentration of solution B = 7.10\times 10^{-6}M7.10×10

−6

M

Explanation:

plz mark as brainlist

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g