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Chemistry: Post your doubts here!

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RoOkaYya G you were right
the answer is 1.3
Here's the solution -->
Given, molar mass of KOH = 56 g mol−1

30% by mass KOH solution means that 30 g of KOH is present in 100 g of the solution.

6.90 M solution contains 6.90 mol of KOH in 1000 cm3 of the solution.

1 mol of KOH = 56 g

6.90 mol of KOH = (56 × 6.90) g

= 386.4 g

∴ 386.4 g of KOH is present =
Chem_2006_XII_SET-I_delhi_richa_LVN_html_5a8485e5.gif


= 1288 g of solution

Density of KOH solution =
Chem_2006_XII_SET-I_delhi_richa_LVN_html_m17fdb72b.gif


Chem_2006_XII_SET-I_delhi_richa_LVN_html_m69699ebc.gif


= 1.288 g cm−3

Hence, the density of the given solution is 1.288 g cm−3


now explain this all plz :p
I didn't get anything :p
 
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RoOkaYya G you were right
the answer is 1.3
Here's the solution -->
Given, molar mass of KOH = 56 g mol−1

30% by mass KOH solution means that 30 g of KOH is present in 100 g of the solution.

6.90 M solution contains 6.90 mol of KOH in 1000 cm3 of the solution.

1 mol of KOH = 56 g

6.90 mol of KOH = (56 × 6.90) g

= 386.4 g

∴ 386.4 g of KOH is present =
Chem_2006_XII_SET-I_delhi_richa_LVN_html_5a8485e5.gif


= 1288 g of solution

Density of KOH solution =
Chem_2006_XII_SET-I_delhi_richa_LVN_html_m17fdb72b.gif


Chem_2006_XII_SET-I_delhi_richa_LVN_html_m69699ebc.gif


= 1.288 g cm−3

Hence, the density of the given solution is 1.288 g cm−3


now explain this all plz :p
I didn't get anything :p
M means mol per dm cube
which means x number of moles in 1000 cm cube of solution

so 6.90 M means tht there is 6.90 mol of of KOH in 1000 cm cube of solution

since there is 30 % by mass of KOH (in pure form) in the total volume of solution
So this means tht theres 0.3 of the whole of the solution only as KOH molecules

since 1 mol of KOH = 56 g
then 6.90 mol as per stated in the question will contain 6.90 * 56 = 386.4 grams of KOH
hence for the KOH in the solution which is in the ratio of 0.3 : 0.7 to water....the mass will be 386.4 / 0.3 = 1288 g in all

since density = mass / volume
then for the 1000 cm cube volume of solution its density will be 1288 / 1000 = 1.28
 

huh

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Hello everyone. How do we predict bond angles and shapes? Advice, notes or anything will be appreciated. I remember in M/J 14, PCl5 bond angle and shape was asked and I had no clue.
 
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Hello everyone. How do we predict bond angles and shapes? Advice, notes or anything will be appreciated. I remember in M/J 14, PCl5 bond angle and shape was asked and I had no clue.
each equatorial P–Cl bond makes two 90° and two 120° bond angles with the other bonds in the molecule
each axial P–Cl bond makes three 90° and one 180° bond angles with the other bonds in the molecule.

it is sp3 hybridised n there is 5 charge clouds (5 bond pair) hence its trigonal bipyramidal
 
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each equatorial P–Cl bond makes two 90° and two 120° bond angles with the other bonds in the molecule
each axial P–Cl bond makes three 90° and one 180° bond angles with the other bonds in the molecule.

it is sp3 hybridised n there is 5 charge clouds (5 bond pair) hence its trigonal bipyramidal
don't confuse it with hybridization P-Cl makes 90 degree with axial and 120 degree with the equatorial
 
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somebody please explain how is the shape of I3^- linear?
doesn't it have 2 bonding and 3 lone?
this is written is syllabus "
*explain the shapes of, and bond angles in, molecules by using
the qualitative model of electron-pair repulsion (including lone
pairs), using as simple examples: BF3 (trigonal), CO2 (linear),
CH4 (tetrahedral), NH3 (pyramidal), H2O (non-linear), SF6 (octahedral),
PF5 (trigonal bipyramid) "
 
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