a. acidic b. basic ; neutral d. will not be affected Find the molarity of 10% NaOH.[MOE 2003 a. 1.5 b! 2.5

Question

a. acidic
b. basic
; neutral
d. will not be affected
Find the molarity of 10% NaOH.[MOE 2003
a. 1.5
b! 2.5
c. 4
d. 0.4
10 ml of 2.5 N NaOH is mixed with 20 ml of
N HCI. The mixture is diluted to 100 ml. WI
is the nature of mixture?
IMOF 20​

in progress 0
Farah 6 months 2021-08-02T15:48:14+00:00 1 Answers 7 views 0

Answers ( )

    0
    2021-08-02T15:49:50+00:00

    1. You didn’t post the question to Number 1.

    2.

    10% means There’s 10g in 100ml of this solution.

    This is the weight/volume(w/v) expression of concentration.

    So

    We have Mass =10g

    volume =100ml

    Molarity = Moles of solute/volume of solution(in LITRES)

    Moles=Mass/Molar mass

    Molar mass of NaOH=40g/mol

    Mole=10/40

    =0.25mole

    Volume =100ml =0.1Litres

    MOLARITY=0.25/0.1 =

    =2.5M

    OPTION B.

    3. NOTE: THE MOLARITY AND NORMALITY OF NAOH AND HCL ARE THE SAME(This doesn’t happen for all compounds tho)

    So We can take 2.5N(Normality) of NaOH to be 2.5M(Molarity) NaOH

    I think you forgot to write the Normality of the second one. I’ll take it to be 1N. Maybe you can then Input supposed value when you’re solving on your own

    So

    1N HCl is same as 1M HCL

    We were given their respective volumes

    2.5M NaOH can also be written as 2.5mole/volume(in Liters)

    The volume of NaOH =10ml or 0.01L

    Moles = 2.5mole/L x 0.01L

    You notice that Liters on top and bottom cancels out… leaving the moles

    So

    Mole=0.01×2.5 = 0.025moles of NaOH

    we’re gonna do the same for 1N HCl in 20ml(0.02L)

    So

    Mole = 1 x 0.02 =0.02moles

    Total Mole = 0.02 + 0.025 =0.045moles

    The Final Volume is 100ml as stated in the question. It was diluted to 100ml or 0.1L

    So

    Final Concentration In Molarity

    = Total Moles/Volume in L

    =0.045/0.1

    =0.45M.

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